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-    "markdown": "---\ntitle: \"Binary response\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse data with a binary outcome?\n-   Can we test if our model is any good?\n-   Be able to perform a logistic regression with a binary outcome\n-   Predict outcomes of new data, based on a defined model\n\n**Objectives**\n\n- Be able to analyse binary outcome data\n- Understand different methods of testing model fit\n- Be able to make model predictions\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n\n# Needed for additional probability functionality\nfrom scipy.stats import *\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/finches_early.csv`\n\nThese data come from an analysis of gene flow across two finch species [@lamichhaney2020]. They are slightly adapted here for illustrative purposes.\n\nThe data focus on two species, _Geospiza fortis_ and _G. scandens_. The original measurements are split by a uniquely timed event: a particularly strong El Niño event in 1983. This event changed the vegetation and food supply of the finches, allowing F1 hybrids of the two species to survive, whereas before 1983 they could not. The measurements are classed as `early` (pre-1983) and `late` (1983 onwards).\n\nHere we are looking only at the `early` data. We are specifically focussing on the beak shape classification, which we saw earlier in @fig-beak_shape_glm.\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nearly_finches <- read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `pointed_beak` column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-6-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `pointed_beak` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a pointed beak (`1`), and those with a blunt one (`0`).\n\nWe can force Python to temporarily covert the data to a factor, by making the `pointed_beak` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-7-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-8-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-9-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea...\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-10-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-11-1.png){width=614}\n:::\n:::\n\n\n:::\n\nOf course this is rubbish - we can't have a beak classification outside the range of $[0, 1]$. It's either blunt (`0`) or pointed (`1`).\n\nBut for the sake of exploration, let's look at the assumptions:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nlm_bks <- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-12-3.png){width=672}\n:::\n:::\n\n\n## Python\n\nFirst, we create a linear model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n```\n:::\n\n\nNext, we can create the diagnostic plots:\n\n::: {.cell}\n\n```{.python .cell-code}\ndgplots(lm_bks_py)\n```\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-15-1.png){width=96}\n:::\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](images/dgplots/2024_01_31-02-26-30_PM_dgplots.png){width=804}\n:::\n:::\n\n\n:::\n\nThey're ~~pretty~~ extremely bad.\n\n-   The response is not linear (Residual Plot, binary response plot, common sense).\n-   The residuals do not appear to be distributed normally (Q-Q Plot)\n-   The variance is not homogeneous across the predicted values (Location-Scale Plot)\n-   But - there is always a silver lining - we don't have influential data points.\n\n## Creating a suitable model\n\nSo far we've established that using a simple linear model to describe a potential relationship between beak length and the probability of having a pointed beak is not a good idea. So, what _can_ we do?\n\nOne of the ways we can deal with binary outcome data is by performing a logistic regression. Instead of fitting a straight line to our data, and performing a regression on that, we fit a line that has an S shape. This avoids the model making predictions outside the $[0, 1]$ range.\n\nWe described our standard linear relationship as follows:\n\n$Y = \\beta_0 + \\beta_1X$\n\nWe can now map this to our non-linear relationship via the **logistic link function**:\n\n$Y = \\frac{\\exp(\\beta_0 + \\beta_1X)}{1 + \\exp(\\beta_0 + \\beta_1X)}$\n\nNote that the $\\beta_0 + \\beta_1X$ part is identical to the formula of a straight line.\n\nThe rest of the function is what makes the straight line curve into its characteristic S shape. \n\n:::{.callout-note collapse=true}\n## Euler's number ($\\exp$): would you like to know more?\n\nIn mathematics, $\\rm e$ represents a constant of around 2.718. Another notation is $\\exp$, which is often used when notations become a bit cumbersome. Here, I exclusively use the $\\exp$ notation for consistency.\n:::\n\n::: {.callout-important}\n## The logistic function\n\nThe shape of the logistic function is hugely influenced by the different parameters, in particular $\\beta_1$. The plots below show different situations, where $\\beta_0 = 0$ in all cases, but $\\beta_1$ varies.\n\nThe first plot shows the logistic function in its simplest form, with the others showing the effect of varying $\\beta_1$.\n\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-17-1.png){width=672}\n:::\n:::\n\n\n* when $\\beta_1 = 1$, this gives the simplest logistic function\n* when $\\beta_1 = 0$ gives a horizontal line, with $Y = \\frac{\\exp(\\beta_0)}{1+\\exp(\\beta_0)}$\n* when $\\beta_1$ is negative flips the curve around, so it slopes down\n* when $\\beta_1$ is very large then the curve becomes extremely steep\n\n:::\n\nWe can fit such an S-shaped curve to our `early_finches` data set, by creating a generalised linear model.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nIn R we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks <- glm(pointed_beak ~ blength,\n               family = binomial,\n               data = early_finches)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\nIf you forget to set the `family` argument, then the `glm()` function will perform a standard linear model fit, identical to what the `lm()` function would do.\n\n## Python\n\nIn Python we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks_py`:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ blength\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_py = model.fit()\n```\n:::\n\n\nThe format of this function is similar to that used by the `ols()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**. This is buried deep inside the `statsmodels` package and needs to be defined as `sm.families.Binomial()`.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_bks)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_bks_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Wed, 31 Jan 2024   Deviance:                       9.1879\nTime:                        14:26:31   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n```\n\n\n:::\n:::\n\n\n:::\n\nThere’s a lot to unpack here, but let's start with what we're familiar with: coefficients!\n\n## Parameter interpretation\n\n::: {.panel-tabset group=\"language\"}\n\n## R\n\nThe coefficients or parameters can be found in the `Coefficients` block. The main numbers to extract from the output are the two numbers underneath `Estimate.Std`:\n\n```\nCoefficients:\n            Estimate Std.\n(Intercept)  -43.410\nblength        3.387 \n```\n\n## Python\n\nRight at the bottom is a table showing the model coefficients. The main numbers to extract from the output are the two numbers in the `coef` column:\n\n```\n======================\n                 coef\n----------------------\nIntercept    -43.4096\nblength        3.3866\n======================\n```\n\n:::\n\nThese are the coefficients of the logistic model equation and need to be placed in the correct equation if we want to be able to calculate the probability of having a pointed beak for a given beak length.\n\nThe $p$ values at the end of each coefficient row merely show whether that particular coefficient is significantly different from zero. This is similar to the $p$ values obtained in the summary output of a linear model. As with continuous predictors in simple models, these $p$ values can be used to decide whether that predictor is important (so in this case beak length appears to be significant). However, these $p$ values aren’t great to work with when we have multiple predictor variables, or when we have categorical predictors with multiple levels (since the output will give us a $p$ value for each level rather than for the predictor as a whole).\n\nWe can use the coefficients to calculate the probability of having a pointed beak for a given beak length:\n\n$$ P(pointed \\ beak) = \\frac{\\exp(-43.41 + 3.39 \\times blength)}{1 + \\exp(-43.41 + 3.39 \\times blength)} $$\n\nHaving this formula means that we can calculate the probability of having a pointed beak for any beak length. How do we work this out in practice? \n\n::: {.panel-tabset group=\"language\"}\n\n## R\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 15) / (1 + exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.9994131\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 10) / (1 + exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 7.410155e-05\n```\n\n\n:::\n:::\n\n\n## Python\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# import the math library\nimport math\n```\n:::\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 15) / (1 + math.exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9994130595039192\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 10) / (1 + math.exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n7.410155028945912e-05\n```\n\n\n:::\n:::\n\n:::\n\nWe can calculate the the probabilities for all our observed values and if we do that then we can see that the larger the beak length is, the higher the probability that a beak shape would be pointed. I'm visualising this together with the logistic curve, where the blue points are the calculated probabilities:\n\n::: {.callout-note collapse=true}\n## Code available here\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment(type.predict = \"response\") %>% \n  ggplot() +\n  geom_point(aes(x = blength, y = pointed_beak)) +\n  geom_line(aes(x = blength, y = .fitted),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = blength, y = .fitted),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py) +\n  geom_point(aes(x = \"blength\", y = \"pointed_beak\")) +\n  geom_line(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\"))\n```\n:::\n\n:::\n:::\n\n\n::: {.cell}\n::: {.cell-output-display}\n![Predicted probabilities for beak classification](glm-practical-logistic-binary_files/figure-html/fig-beak_class_glm_probs-2.png){#fig-beak_class_glm_probs width=672}\n:::\n:::\n\n\nThe graph shows us that, based on the data that we have and the model we used to make predictions about our response variable, the probability of seeing a pointed beak increases with beak length.\n\nShort beaks are more closely associated with the bluntly shaped beaks, whereas long beaks are more closely associated with the pointed shape. It's also clear that there is a range of beak lengths (around 13 mm) where the probability of getting one shape or another is much more even.\n\n<!-- ## Parameter estimation explained -->\n\n<!-- Now that we know how to interpret the coefficients or parameters, let's have a look at how they're actually determined. To understand this, we need to take a step back. -->\n\n<!-- * Sum of squares (OLS) -->\n<!-- * MLE -->\n<!-- * Deviance -->\n\n<!-- ## Assumptions -->\n\n<!-- * GAMLSS -->\n\n\n## Assumptions\n\nAs explained in the background chapter, we can't really use the standard diagnostic plots to assess assumptions. We're not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_bks , which = 4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-30-1.png){width=672}\n:::\n:::\n\n\n:::{.callout-note collapse=true}\n## Extracting the Cook's distances from the `glm` object\n\nInstead of using the `plot()` function, we can also extract the values directly from the `glm` object. We can use the `augment()` function to do this and create a lollipop or stem plot:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment() %>%            # get underlying data\n  select(.cooksd) %>%      # select the Cook's d\n  mutate(obs = 1:n()) %>%  # create an index column\n  ggplot(aes(x = obs, y = .cooksd)) +\n  geom_point() +\n  geom_segment(aes(x = obs, y = .cooksd, xend = obs, yend = 0))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-31-1.png){width=672}\n:::\n:::\n\n:::\n\n## Python\n\nAs always, there are different ways of doing this. Here we extract the Cook's d values from the `glm` object and put them in a Pandas DataFrame. We can then use that to plot them in a lollipop or stem plot.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# extract the Cook's distances\nglm_bks_py_resid = pd.DataFrame(glm_bks_py.\n                                get_influence().\n                                summary_frame()[\"cooks_d\"])\n\n# add row index \nglm_bks_py_resid['obs'] = glm_bks_py_resid.reset_index().index\n```\n:::\n\n\nWe now have two columns:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nglm_bks_py_resid\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n         cooks_d  obs\n0   1.854360e-07    0\n1   3.388262e-07    1\n2   3.217960e-05    2\n3   1.194847e-05    3\n4   6.643975e-06    4\n..           ...  ...\n56  1.225519e-05   56\n57  2.484468e-05   57\n58  6.781364e-06   58\n59  1.850240e-07   59\n60  3.532360e-05   60\n\n[61 rows x 2 columns]\n```\n\n\n:::\n:::\n\n\nWe can use these to create the plot:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(glm_bks_py_resid,\n         aes(x = \"obs\",\n             y = \"cooks_d\")) +\n     geom_segment(aes(x = \"obs\",y = \"cooks_d\", xend = \"obs\", yend = 0)) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-34-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis shows that there are no very obvious influential points. You could regard point `34` as potentially influential (it's got a Cook's distance of around `0.8`), but I'm not overly worried.\n\nIf we were worried, we'd remove the troublesome data point, re-run the analysis and see if that changes the statistical outcome. If so, then our entire (statistical) conclusion hinges on one data point, which is not a very robust bit of research. If it *doesn't* change our significance, then all is well, even though that data point is influential.\n\n## Assessing significance\n\nWe can ask several questions.\n\n**Is the model well-specified?**\n\nRoughly speaking this asks \"can our model predict our data reasonably well?\"\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves. \n\n\n::: {.cell}\n\n```{.r .cell-code}\npchisq(9.1879, 59, lower.tail = FALSE)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1\n```\n\n\n:::\n:::\n\n\n\nHere, we’ve used the `pchisq` function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nfrom scipy.stats import chi2\n\nchi2.sf(9.1879, 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9999999999999916\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives us a probability of `1`. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\n\n**Is the overall model better than the null model?**\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\npchisq(84.5476 - 9.1879, 60 - 59, lower.tail = FALSE)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 3.923163e-18\n```\n\n\n:::\n:::\n\n\nHere we’ve used the `pchisq` function again (if you didn’t ask before, you probably aren’t going to ask now).\n\n## Python\n\nFirst we need to define the null model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ 1\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_null_py = model.fit()\n\nprint(glm_bks_null_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       60\nModel Family:                Binomial   Df Model:                            0\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -42.274\nDate:                Wed, 31 Jan 2024   Deviance:                       84.548\nTime:                        14:26:35   Pearson chi2:                     61.0\nNo. Iterations:                     3   Pseudo R-squ. (CS):              0.000\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept      0.0328      0.256      0.128      0.898      -0.469       0.535\n==============================================================================\n```\n\n\n:::\n:::\n\n\nIn order to compare our original fitted model to the null model we need to know the deviances of both models and the residual degrees of freedom of both models, which we could get from the summary method.\n\n\n::: {.cell}\n\n```{.python .cell-code}\nchi2.sf(84.5476 - 9.1879, 60 - 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n3.9231627082752525e-18\n```\n\n\n:::\n:::\n\n\n:::\n\nThe first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\n\nThis gives us a probability of pretty much zero. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with `blength` as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\n\n**Are any of the individual predictors significant?**\n\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the beak length predictor is significant.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_bks, test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe `anova()` function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\n\nThe p-value for the `blength` predictor is written under then Pr(>Chi) column and we can see that it is less than `< 2.2e-16`. So, beak length is a significant predictor.\n\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n## Python\n\nAlas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, not beknownst to me...\n\n:::\n\n\n\n\n## Exercises\n\n### Diabetes {#sec-exr_diabetes}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/diabetes.csv`.\n\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called `test_result`, `glucose` and `diastolic`.\n\nWe want to see if the `glucose` tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\n1. Load and visualise the data\n2. Create a suitable model\n3. Determine if there are any statistically significant predictors\n4. Calculate the probability of a positive diabetes test result for a glucose tolerance test value of `glucose = 150`\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\ndiabetes <- read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `test_result` column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data, by outcome:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-43-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `test_result` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\n\nWe can force Python to temporarily covert the data to a factor, by making the `test_result` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-44-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-45-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-46-1.png){width=614}\n:::\n:::\n\n\n:::\n\n#### Create a suitable model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe'll use the `glm()` function to create a generalised linear model. Here we save the model in an object called `glm_dia`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_dia <- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n```\n:::\n\n\n:::\n\nLet's look at the model parameters:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_dia)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) -5.611732   0.442289  -12.69   <2e-16 ***\nglucose      0.039510   0.003398   11.63   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_dia_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Wed, 31 Jan 2024   Deviance:                       752.20\nTime:                        14:26:37   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n```\n\n\n:::\n:::\n\n:::\n\nWe can see that `glucose` is a significant predictor for the `test_result` (the $p$ value is much smaller than 0.05).\n\nKnowing this, we're interested in the coefficients. We have an intercept of `-5.61` and `0.0395` for `glucose`. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\n$$ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} $$\n\n#### Calculating probabilities\n\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given `glucose` value. If we do this for `glucose = 150`, we get the following:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5962827\n```\n\n\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.5962826992967878\n```\n\n\n:::\n:::\n\n:::\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We use a logistic regression to model a binary response\n-   We can feed new observations into the model and get probabilities for the outcome\n:::\n",
+    "markdown": "---\ntitle: \"Binary response\"\n---\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n\n:::\n\n\n::: {.callout-tip}\n## Learning outcomes\n\n**Questions**\n\n-   How do we analyse data with a binary outcome?\n-   Can we test if our model is any good?\n-   Be able to perform a logistic regression with a binary outcome\n-   Predict outcomes of new data, based on a defined model\n\n**Objectives**\n\n- Be able to analyse binary outcome data\n- Understand different methods of testing model fit\n- Be able to make model predictions\n:::\n\n## Libraries and functions\n\n::: {.callout-note collapse=\"true\"}\n## Click to expand\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n### Libraries\n### Functions\n\n## Python\n\n### Libraries\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# A maths library\nimport math\n# A Python data analysis and manipulation tool\nimport pandas as pd\n\n# Python equivalent of `ggplot2`\nfrom plotnine import *\n\n# Statistical models, conducting tests and statistical data exploration\nimport statsmodels.api as sm\n\n# Convenience interface for specifying models using formula strings and DataFrames\nimport statsmodels.formula.api as smf\n\n# Needed for additional probability functionality\nfrom scipy.stats import *\n```\n:::\n\n\n### Functions\n:::\n:::\n\nThe example in this section uses the following data set:\n\n`data/finches_early.csv`\n\nThese data come from an analysis of gene flow across two finch species [@lamichhaney2020]. They are slightly adapted here for illustrative purposes.\n\nThe data focus on two species, _Geospiza fortis_ and _G. scandens_. The original measurements are split by a uniquely timed event: a particularly strong El Niño event in 1983. This event changed the vegetation and food supply of the finches, allowing F1 hybrids of the two species to survive, whereas before 1983 they could not. The measurements are classed as `early` (pre-1983) and `late` (1983 onwards).\n\nHere we are looking only at the `early` data. We are specifically focussing on the beak shape classification, which we saw earlier in @fig-beak_shape_glm.\n\n## Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nearly_finches <- read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `pointed_beak` column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-6-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `pointed_beak` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a pointed beak (`1`), and those with a blunt one (`0`).\n\nWe can force Python to temporarily covert the data to a factor, by making the `pointed_beak` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-7-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-8-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-9-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea...\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-10-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-11-1.png){width=614}\n:::\n:::\n\n\n:::\n\nOf course this is rubbish - we can't have a beak classification outside the range of $[0, 1]$. It's either blunt (`0`) or pointed (`1`).\n\nBut for the sake of exploration, let's look at the assumptions:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nlm_bks <- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-12-3.png){width=672}\n:::\n:::\n\n\n## Python\n\nFirst, we create a linear model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n```\n:::\n\n\nNext, we can create the diagnostic plots:\n\n::: {.cell}\n\n```{.python .cell-code}\ndgplots(lm_bks_py)\n```\n:::\n\n::: {.cell}\n\n:::\n\n::: {.cell}\n::: {.cell-output-display}\n![](images/dgplots/2024_02_01-07-29-38_AM_dgplots.png){width=805}\n:::\n:::\n\n\n:::\n\nThey're ~~pretty~~ extremely bad.\n\n-   The response is not linear (Residual Plot, binary response plot, common sense).\n-   The residuals do not appear to be distributed normally (Q-Q Plot)\n-   The variance is not homogeneous across the predicted values (Location-Scale Plot)\n-   But - there is always a silver lining - we don't have influential data points.\n\n## Creating a suitable model\n\nSo far we've established that using a simple linear model to describe a potential relationship between beak length and the probability of having a pointed beak is not a good idea. So, what _can_ we do?\n\nOne of the ways we can deal with binary outcome data is by performing a logistic regression. Instead of fitting a straight line to our data, and performing a regression on that, we fit a line that has an S shape. This avoids the model making predictions outside the $[0, 1]$ range.\n\nWe described our standard linear relationship as follows:\n\n$Y = \\beta_0 + \\beta_1X$\n\nWe can now map this to our non-linear relationship via the **logistic link function**:\n\n$Y = \\frac{\\exp(\\beta_0 + \\beta_1X)}{1 + \\exp(\\beta_0 + \\beta_1X)}$\n\nNote that the $\\beta_0 + \\beta_1X$ part is identical to the formula of a straight line.\n\nThe rest of the function is what makes the straight line curve into its characteristic S shape. \n\n:::{.callout-note collapse=true}\n## Euler's number ($\\exp$): would you like to know more?\n\nIn mathematics, $\\rm e$ represents a constant of around 2.718. Another notation is $\\exp$, which is often used when notations become a bit cumbersome. Here, I exclusively use the $\\exp$ notation for consistency.\n:::\n\n::: {.callout-important}\n## The logistic function\n\nThe shape of the logistic function is hugely influenced by the different parameters, in particular $\\beta_1$. The plots below show different situations, where $\\beta_0 = 0$ in all cases, but $\\beta_1$ varies.\n\nThe first plot shows the logistic function in its simplest form, with the others showing the effect of varying $\\beta_1$.\n\n\n::: {.cell}\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-17-1.png){width=672}\n:::\n:::\n\n\n* when $\\beta_1 = 1$, this gives the simplest logistic function\n* when $\\beta_1 = 0$ gives a horizontal line, with $Y = \\frac{\\exp(\\beta_0)}{1+\\exp(\\beta_0)}$\n* when $\\beta_1$ is negative flips the curve around, so it slopes down\n* when $\\beta_1$ is very large then the curve becomes extremely steep\n\n:::\n\nWe can fit such an S-shaped curve to our `early_finches` data set, by creating a generalised linear model.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nIn R we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks <- glm(pointed_beak ~ blength,\n               family = binomial,\n               data = early_finches)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\nIf you forget to set the `family` argument, then the `glm()` function will perform a standard linear model fit, identical to what the `lm()` function would do.\n\n## Python\n\nIn Python we have a few options to do this, and by far the most familiar function would be `glm()`. Here we save the model in an object called `glm_bks_py`:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ blength\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_py = model.fit()\n```\n:::\n\n\nThe format of this function is similar to that used by the `ols()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**. This is buried deep inside the `statsmodels` package and needs to be defined as `sm.families.Binomial()`.\n\n:::\n\n## Model output\n\nThat's the easy part done! The trickier part is interpreting the output. First of all, we'll get some summary information.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_bks)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(>|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_bks_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Thu, 01 Feb 2024   Deviance:                       9.1879\nTime:                        07:29:39   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n```\n\n\n:::\n:::\n\n\n:::\n\nThere’s a lot to unpack here, but let's start with what we're familiar with: coefficients!\n\n## Parameter interpretation\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nThe coefficients or parameters can be found in the `Coefficients` block. The main numbers to extract from the output are the two numbers underneath `Estimate.Std`:\n\n```\nCoefficients:\n            Estimate Std.\n(Intercept)  -43.410\nblength        3.387 \n```\n\n## Python\n\nRight at the bottom is a table showing the model coefficients. The main numbers to extract from the output are the two numbers in the `coef` column:\n\n```\n======================\n                 coef\n----------------------\nIntercept    -43.4096\nblength        3.3866\n======================\n```\n\n:::\n\nThese are the coefficients of the logistic model equation and need to be placed in the correct equation if we want to be able to calculate the probability of having a pointed beak for a given beak length.\n\nThe $p$ values at the end of each coefficient row merely show whether that particular coefficient is significantly different from zero. This is similar to the $p$ values obtained in the summary output of a linear model. As with continuous predictors in simple models, these $p$ values can be used to decide whether that predictor is important (so in this case beak length appears to be significant). However, these $p$ values aren’t great to work with when we have multiple predictor variables, or when we have categorical predictors with multiple levels (since the output will give us a $p$ value for each level rather than for the predictor as a whole).\n\nWe can use the coefficients to calculate the probability of having a pointed beak for a given beak length:\n\n$$ P(pointed \\ beak) = \\frac{\\exp(-43.41 + 3.39 \\times blength)}{1 + \\exp(-43.41 + 3.39 \\times blength)} $$\n\nHaving this formula means that we can calculate the probability of having a pointed beak for any beak length. How do we work this out in practice? \n\n::: {.panel-tabset group=\"language\"}\n\n## R\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 15) / (1 + exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.9994131\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-43.41 + 3.39 * 10) / (1 + exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 7.410155e-05\n```\n\n\n:::\n:::\n\n\n## Python\n\nWell, the probability of having a pointed beak if the beak length is large (for example 15 mm) can be calculated as follows:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# import the math library\nimport math\n```\n:::\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 15) / (1 + math.exp(-43.41 + 3.39 * 15))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9994130595039192\n```\n\n\n:::\n:::\n\n\nIf the beak length is small (for example 10 mm), the probability of having a pointed beak is extremely low:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-43.41 + 3.39 * 10) / (1 + math.exp(-43.41 + 3.39 * 10))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n7.410155028945912e-05\n```\n\n\n:::\n:::\n\n:::\n\nWe can calculate the the probabilities for all our observed values and if we do that then we can see that the larger the beak length is, the higher the probability that a beak shape would be pointed. I'm visualising this together with the logistic curve, where the blue points are the calculated probabilities:\n\n::: {.callout-note collapse=true}\n## Code available here\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment(type.predict = \"response\") %>% \n  ggplot() +\n  geom_point(aes(x = blength, y = pointed_beak)) +\n  geom_line(aes(x = blength, y = .fitted),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = blength, y = .fitted),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(early_finches_py) +\n  geom_point(aes(x = \"blength\", y = \"pointed_beak\")) +\n  geom_line(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n            linetype = \"dashed\",\n            colour = \"blue\") +\n  geom_point(aes(x = \"blength\", y = glm_bks_py.fittedvalues),\n             colour = \"blue\", alpha = 0.5) +\n  labs(x = \"beak length (mm)\",\n       y = \"Probability\"))\n```\n:::\n\n:::\n:::\n\n\n::: {.cell}\n::: {.cell-output-display}\n![Predicted probabilities for beak classification](glm-practical-logistic-binary_files/figure-html/fig-beak_class_glm_probs-2.png){#fig-beak_class_glm_probs width=672}\n:::\n:::\n\n\nThe graph shows us that, based on the data that we have and the model we used to make predictions about our response variable, the probability of seeing a pointed beak increases with beak length.\n\nShort beaks are more closely associated with the bluntly shaped beaks, whereas long beaks are more closely associated with the pointed shape. It's also clear that there is a range of beak lengths (around 13 mm) where the probability of getting one shape or another is much more even.\n\n<!-- ## Parameter estimation explained -->\n\n<!-- Now that we know how to interpret the coefficients or parameters, let's have a look at how they're actually determined. To understand this, we need to take a step back. -->\n\n<!-- * Sum of squares (OLS) -->\n<!-- * MLE -->\n<!-- * Deviance -->\n\n<!-- ## Assumptions -->\n\n<!-- * GAMLSS -->\n\n\n## Assumptions\n\nAs explained in the background chapter, we can't really use the standard diagnostic plots to assess assumptions. We're not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nplot(glm_bks , which = 4)\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-30-1.png){width=672}\n:::\n:::\n\n\n:::{.callout-note collapse=true}\n## Extracting the Cook's distances from the `glm` object\n\nInstead of using the `plot()` function, we can also extract the values directly from the `glm` object. We can use the `augment()` function to do this and create a lollipop or stem plot:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_bks %>% \n  augment() %>%            # get underlying data\n  select(.cooksd) %>%      # select the Cook's d\n  mutate(obs = 1:n()) %>%  # create an index column\n  ggplot(aes(x = obs, y = .cooksd)) +\n  geom_point() +\n  geom_segment(aes(x = obs, y = .cooksd, xend = obs, yend = 0))\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-31-1.png){width=672}\n:::\n:::\n\n:::\n\n## Python\n\nAs always, there are different ways of doing this. Here we extract the Cook's d values from the `glm` object and put them in a Pandas DataFrame. We can then use that to plot them in a lollipop or stem plot.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# extract the Cook's distances\nglm_bks_py_resid = pd.DataFrame(glm_bks_py.\n                                get_influence().\n                                summary_frame()[\"cooks_d\"])\n\n# add row index \nglm_bks_py_resid['obs'] = glm_bks_py_resid.reset_index().index\n```\n:::\n\n\nWe now have two columns:\n\n\n::: {.cell}\n\n```{.python .cell-code}\nglm_bks_py_resid\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n         cooks_d  obs\n0   1.854360e-07    0\n1   3.388262e-07    1\n2   3.217960e-05    2\n3   1.194847e-05    3\n4   6.643975e-06    4\n..           ...  ...\n56  1.225519e-05   56\n57  2.484468e-05   57\n58  6.781364e-06   58\n59  1.850240e-07   59\n60  3.532360e-05   60\n\n[61 rows x 2 columns]\n```\n\n\n:::\n:::\n\n\nWe can use these to create the plot:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(glm_bks_py_resid,\n         aes(x = \"obs\",\n             y = \"cooks_d\")) +\n     geom_segment(aes(x = \"obs\", y = \"cooks_d\", xend = \"obs\", yend = 0)) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-34-1.png){width=614}\n:::\n:::\n\n\n:::\n\nThis shows that there are no very obvious influential points. You could regard point `34` as potentially influential (it's got a Cook's distance of around `0.8`), but I'm not overly worried.\n\nIf we were worried, we'd remove the troublesome data point, re-run the analysis and see if that changes the statistical outcome. If so, then our entire (statistical) conclusion hinges on one data point, which is not a very robust bit of research. If it *doesn't* change our significance, then all is well, even though that data point is influential.\n\n## Assessing significance\n\nWe can ask several questions.\n\n**Is the model well-specified?**\n\nRoughly speaking this asks \"can our model predict our data reasonably well?\"\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves. \n\n\n::: {.cell}\n\n```{.r .cell-code}\npchisq(9.1879, 59, lower.tail = FALSE)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 1\n```\n\n\n:::\n:::\n\n\n\nHere, we’ve used the `pchisq` function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nfrom scipy.stats import chi2\n\nchi2.sf(9.1879, 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.9999999999999916\n```\n\n\n:::\n:::\n\n\n:::\n\nThis gives us a probability of `1`. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\n\n**Is the overall model better than the null model?**\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\npchisq(84.5476 - 9.1879, 60 - 59, lower.tail = FALSE)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 3.923163e-18\n```\n\n\n:::\n:::\n\n\nHere we’ve used the `pchisq` function again (if you didn’t ask before, you probably aren’t going to ask now).\n\n## Python\n\nFirst we need to define the null model:\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ 1\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_null_py = model.fit()\n\nprint(glm_bks_null_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       60\nModel Family:                Binomial   Df Model:                            0\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -42.274\nDate:                Thu, 01 Feb 2024   Deviance:                       84.548\nTime:                        07:29:43   Pearson chi2:                     61.0\nNo. Iterations:                     3   Pseudo R-squ. (CS):              0.000\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept      0.0328      0.256      0.128      0.898      -0.469       0.535\n==============================================================================\n```\n\n\n:::\n:::\n\n\nIn order to compare our original fitted model to the null model we need to know the deviances of both models and the residual degrees of freedom of both models, which we could get from the summary method.\n\n\n::: {.cell}\n\n```{.python .cell-code}\nchi2.sf(84.5476 - 9.1879, 60 - 59)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n3.9231627082752525e-18\n```\n\n\n:::\n:::\n\n\n:::\n\nThe first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\n\nThis gives us a probability of pretty much zero. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with `blength` as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\n\n**Are any of the individual predictors significant?**\n\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the beak length predictor is significant.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nanova(glm_bks, test = \"Chisq\")\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 < 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n```\n\n\n:::\n:::\n\n\nThe `anova()` function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\n\nThe p-value for the `blength` predictor is written under then Pr(>Chi) column and we can see that it is less than `< 2.2e-16`. So, beak length is a significant predictor.\n\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n## Python\n\nAlas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, unbeknownst to me...\n\n:::\n\n\n## Exercises\n\n### Diabetes {#sec-exr_diabetes}\n\n:::{.callout-exercise}\n\n\n{{< level 2 >}}\n\n\n\nFor this exercise we'll be using the data from `data/diabetes.csv`.\n\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called `test_result`, `glucose` and `diastolic`.\n\nWe want to see if the `glucose` tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\n1. Load and visualise the data\n2. Create a suitable model\n3. Determine if there are any statistically significant predictors\n4. Calculate the probability of a positive diabetes test result for a glucose tolerance test value of `glucose = 150`\n\n::: {.callout-answer collapse=\"true\"}\n\n#### Load and visualise the data\n\nFirst we load the data, then we visualise it.\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\ndiabetes <- read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n```\n:::\n\n\n:::\n\nLooking at the data, we can see that the `test_result` column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\n\nWe'll have to deal with this soon. For now, we can plot the data, by outcome:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-43-1.png){width=672}\n:::\n:::\n\n\n## Python\n\nWe could just give Python the `test_result` data directly, but then it would view the values as numeric. Which doesn't really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\n\nWe can force Python to temporarily covert the data to a factor, by making the `test_result` column an `object` type. We can do this directly inside the `ggplot()` function.\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-44-1.png){width=614}\n:::\n:::\n\n:::\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\n\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-45-3.png){width=672}\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n```\n\n::: {.cell-output-display}\n![](glm-practical-logistic-binary_files/figure-html/unnamed-chunk-46-1.png){width=614}\n:::\n:::\n\n\n:::\n\n#### Create a suitable model\n\n::: {.panel-tabset group=\"language\"}\n## R\n\nWe'll use the `glm()` function to create a generalised linear model. Here we save the model in an object called `glm_dia`:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nglm_dia <- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n```\n:::\n\n\nThe format of this function is similar to that used by the `lm()` function for linear models. The important difference is that we must specify the _family_ of error distribution to use. For logistic regression we must set the family to **binomial**.\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n```\n:::\n\n\n:::\n\nLet's look at the model parameters:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nsummary(glm_dia)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(>|z|)    \n(Intercept) -5.611732   0.442289  -12.69   <2e-16 ***\nglucose      0.039510   0.003398   11.63   <2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n```\n\n\n:::\n:::\n\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nprint(glm_dia_py.summary())\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Thu, 01 Feb 2024   Deviance:                       752.20\nTime:                        07:29:45   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P>|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n```\n\n\n:::\n:::\n\n:::\n\nWe can see that `glucose` is a significant predictor for the `test_result` (the $p$ value is much smaller than 0.05).\n\nKnowing this, we're interested in the coefficients. We have an intercept of `-5.61` and `0.0395` for `glucose`. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\n$$ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} $$\n\n#### Calculating probabilities\n\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given `glucose` value. If we do this for `glucose = 150`, we get the following:\n\n::: {.panel-tabset group=\"language\"}\n## R\n\n\n::: {.cell}\n\n```{.r .cell-code}\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5962827\n```\n\n\n:::\n:::\n\n\n## Python\n\n\n::: {.cell}\n\n```{.python .cell-code}\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n0.5962826992967878\n```\n\n\n:::\n:::\n\n:::\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.\n\n:::\n:::\n\n## Summary\n\n::: {.callout-tip}\n#### Key points\n\n-   We use a logistic regression to model a binary response\n-   We can feed new observations into the model and get probabilities for the outcome\n:::\n",
     "supporting": [
       "glm-practical-logistic-binary_files"
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@@ -598,15 +598,7 @@ <h1 class="title">
 <div class="cell">
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-<figure class="figure"><p><img src="glm-practical-logistic-binary_files/figure-html/unnamed-chunk-15-1.png" class="img-fluid figure-img" width="96"></p>
-</figure>
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-<div>
-<figure class="figure"><p><img src="images/dgplots/2024_01_31-02-26-30_PM_dgplots.png" class="img-fluid figure-img" width="804"></p>
+<figure class="figure"><p><img src="images/dgplots/2024_02_01-07-29-38_AM_dgplots.png" class="img-fluid figure-img" width="805"></p>
 </figure>
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@@ -751,8 +743,8 @@ <h1 class="title">
 Model Family:                Binomial   Df Model:                            1
 Link Function:                  Logit   Scale:                          1.0000
 Method:                          IRLS   Log-Likelihood:                -4.5939
-Date:                Wed, 31 Jan 2024   Deviance:                       9.1879
-Time:                        14:26:31   Pearson chi2:                     15.1
+Date:                Thu, 01 Feb 2024   Deviance:                       9.1879
+Time:                        07:29:39   Pearson chi2:                     15.1
 No. Iterations:                     8   Pseudo R-squ. (CS):             0.7093
 Covariance Type:            nonrobust                                         
 ==============================================================================
@@ -1002,7 +994,7 @@ <h1 class="title">
 <div class="sourceCode cell-code" id="cb37"><pre class="sourceCode python code-with-copy"><code class="sourceCode python"><span id="cb37-1"><a href="#cb37-1" aria-hidden="true" tabindex="-1"></a>(ggplot(glm_bks_py_resid,</span>
 <span id="cb37-2"><a href="#cb37-2" aria-hidden="true" tabindex="-1"></a>         aes(x <span class="op">=</span> <span class="st">"obs"</span>,</span>
 <span id="cb37-3"><a href="#cb37-3" aria-hidden="true" tabindex="-1"></a>             y <span class="op">=</span> <span class="st">"cooks_d"</span>)) <span class="op">+</span></span>
-<span id="cb37-4"><a href="#cb37-4" aria-hidden="true" tabindex="-1"></a>     geom_segment(aes(x <span class="op">=</span> <span class="st">"obs"</span>,y <span class="op">=</span> <span class="st">"cooks_d"</span>, xend <span class="op">=</span> <span class="st">"obs"</span>, yend <span class="op">=</span> <span class="dv">0</span>)) <span class="op">+</span></span>
+<span id="cb37-4"><a href="#cb37-4" aria-hidden="true" tabindex="-1"></a>     geom_segment(aes(x <span class="op">=</span> <span class="st">"obs"</span>, y <span class="op">=</span> <span class="st">"cooks_d"</span>, xend <span class="op">=</span> <span class="st">"obs"</span>, yend <span class="op">=</span> <span class="dv">0</span>)) <span class="op">+</span></span>
 <span id="cb37-5"><a href="#cb37-5" aria-hidden="true" tabindex="-1"></a>     geom_point())</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
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@@ -1085,8 +1077,8 @@ <h1 class="title">
 Model Family:                Binomial   Df Model:                            0
 Link Function:                  Logit   Scale:                          1.0000
 Method:                          IRLS   Log-Likelihood:                -42.274
-Date:                Wed, 31 Jan 2024   Deviance:                       84.548
-Time:                        14:26:35   Pearson chi2:                     61.0
+Date:                Thu, 01 Feb 2024   Deviance:                       84.548
+Time:                        07:29:43   Pearson chi2:                     61.0
 No. Iterations:                     3   Pseudo R-squ. (CS):              0.000
 Covariance Type:            nonrobust                                         
 ==============================================================================
@@ -1140,7 +1132,7 @@ <h1 class="title">
 <p>This shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.</p>
 </div>
 <div id="tabset-15-2" class="tab-pane" role="tabpanel" aria-labelledby="tabset-15-2-tab">
-<p>Alas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, not beknownst to me…</p>
+<p>Alas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, unbeknownst to me…</p>
 </div>
 </div>
 </div>
@@ -1356,8 +1348,8 @@ <h1 class="title">
 Model Family:                Binomial   Df Model:                            1
 Link Function:                  Logit   Scale:                          1.0000
 Method:                          IRLS   Log-Likelihood:                -376.10
-Date:                Wed, 31 Jan 2024   Deviance:                       752.20
-Time:                        14:26:37   Pearson chi2:                     713.
+Date:                Thu, 01 Feb 2024   Deviance:                       752.20
+Time:                        07:29:45   Pearson chi2:                     713.
 No. Iterations:                     4   Pseudo R-squ. (CS):             0.2238
 Covariance Type:            nonrobust                                         
 ==============================================================================
diff --git a/_site/search.json b/_site/search.json
index a64a3ee..d38010e 100644
--- a/_site/search.json
+++ b/_site/search.json
@@ -191,7 +191,7 @@
     "href": "materials/glm-practical-logistic-binary.html#load-and-visualise-the-data",
     "title": "\n7  Binary response\n",
     "section": "\n7.2 Load and visualise the data",
-    "text": "7.2 Load and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\nearly_finches &lt;- read_csv(\"data/finches_early.csv\")\n\n\n\n\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n\n\n\n\nLooking at the data, we can see that the pointed_beak column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data:\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the pointed_beak data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a pointed beak (1), and those with a blunt one (0).\nWe can force Python to temporarily covert the data to a factor, by making the pointed_beak column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea…\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n\n\n\n\n\n\n\n\n\n\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n\n\n\n\n\n\n\n\n\n\nOf course this is rubbish - we can’t have a beak classification outside the range of \\([0, 1]\\). It’s either blunt (0) or pointed (1).\nBut for the sake of exploration, let’s look at the assumptions:\n\n\nR\nPython\n\n\n\n\nlm_bks &lt;- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n\n\n\n\n\n\n\n\n\nFirst, we create a linear model:\n\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n\nNext, we can create the diagnostic plots:\n\ndgplots(lm_bks_py)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nThey’re pretty extremely bad.\n\nThe response is not linear (Residual Plot, binary response plot, common sense).\nThe residuals do not appear to be distributed normally (Q-Q Plot)\nThe variance is not homogeneous across the predicted values (Location-Scale Plot)\nBut - there is always a silver lining - we don’t have influential data points.",
+    "text": "7.2 Load and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\nearly_finches &lt;- read_csv(\"data/finches_early.csv\")\n\n\n\n\nearly_finches_py = pd.read_csv(\"data/finches_early.csv\")\n\n\n\n\nLooking at the data, we can see that the pointed_beak column contains zeros and ones. These are actually yes/no classification outcomes and not numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data:\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = factor(pointed_beak),\n          y = blength)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the pointed_beak data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a pointed beak (1), and those with a blunt one (0).\nWe can force Python to temporarily covert the data to a factor, by making the pointed_beak column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(early_finches_py,\n         aes(x = early_finches_py.pointed_beak.astype(object),\n             y = \"blength\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the finches with blunt beaks generally have shorter beak lengths.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nThis presents us with a bit of an issue. We could fit a linear regression model to these data, although we already know that this is a bad idea…\n\n\nR\nPython\n\n\n\n\nggplot(early_finches,\n       aes(x = blength, y = pointed_beak)) +\n  geom_point() +\n  geom_smooth(method = \"lm\", se = FALSE)\n\n\n\n\n\n\n\n\n\n\n(ggplot(early_finches_py,\n         aes(x = \"blength\",\n             y = \"pointed_beak\")) +\n     geom_point() +\n     geom_smooth(method = \"lm\",\n                 colour = \"blue\",\n                 se = False))\n\n\n\n\n\n\n\n\n\n\nOf course this is rubbish - we can’t have a beak classification outside the range of \\([0, 1]\\). It’s either blunt (0) or pointed (1).\nBut for the sake of exploration, let’s look at the assumptions:\n\n\nR\nPython\n\n\n\n\nlm_bks &lt;- lm(pointed_beak ~ blength,\n             data = early_finches)\n\nresid_panel(lm_bks,\n            plots = c(\"resid\", \"qq\", \"ls\", \"cookd\"),\n            smoother = TRUE)\n\n\n\n\n\n\n\n\n\nFirst, we create a linear model:\n\n# create a linear model\nmodel = smf.ols(formula = \"pointed_beak ~ blength\",\n                data = early_finches_py)\n# and get the fitted parameters of the model\nlm_bks_py = model.fit()\n\nNext, we can create the diagnostic plots:\n\ndgplots(lm_bks_py)\n\n\n\n\n\n\n\n\n\n\n\n\nThey’re pretty extremely bad.\n\nThe response is not linear (Residual Plot, binary response plot, common sense).\nThe residuals do not appear to be distributed normally (Q-Q Plot)\nThe variance is not homogeneous across the predicted values (Location-Scale Plot)\nBut - there is always a silver lining - we don’t have influential data points.",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -213,7 +213,7 @@
     "href": "materials/glm-practical-logistic-binary.html#model-output",
     "title": "\n7  Binary response\n",
     "section": "\n7.4 Model output",
-    "text": "7.4 Model output\nThat’s the easy part done! The trickier part is interpreting the output. First of all, we’ll get some summary information.\n\n\nR\nPython\n\n\n\n\nsummary(glm_bks)\n\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n\n\n\n\n\nprint(glm_bks_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Wed, 31 Jan 2024   Deviance:                       9.1879\nTime:                        14:26:31   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n\n\n\n\n\nThere’s a lot to unpack here, but let’s start with what we’re familiar with: coefficients!",
+    "text": "7.4 Model output\nThat’s the easy part done! The trickier part is interpreting the output. First of all, we’ll get some summary information.\n\n\nR\nPython\n\n\n\n\nsummary(glm_bks)\n\n\nCall:\nglm(formula = pointed_beak ~ blength, family = binomial, data = early_finches)\n\nCoefficients:\n            Estimate Std. Error z value Pr(&gt;|z|)   \n(Intercept)  -43.410     15.250  -2.847  0.00442 **\nblength        3.387      1.193   2.839  0.00452 **\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 84.5476  on 60  degrees of freedom\nResidual deviance:  9.1879  on 59  degrees of freedom\nAIC: 13.188\n\nNumber of Fisher Scoring iterations: 8\n\n\n\n\n\nprint(glm_bks_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       59\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -4.5939\nDate:                Thu, 01 Feb 2024   Deviance:                       9.1879\nTime:                        07:29:39   Pearson chi2:                     15.1\nNo. Iterations:                     8   Pseudo R-squ. (CS):             0.7093\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept    -43.4096     15.250     -2.847      0.004     -73.298     -13.521\nblength        3.3866      1.193      2.839      0.005       1.049       5.724\n==============================================================================\n\n\n\n\n\nThere’s a lot to unpack here, but let’s start with what we’re familiar with: coefficients!",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -235,7 +235,7 @@
     "href": "materials/glm-practical-logistic-binary.html#assumptions",
     "title": "\n7  Binary response\n",
     "section": "\n7.6 Assumptions",
-    "text": "7.6 Assumptions\nAs explained in the background chapter, we can’t really use the standard diagnostic plots to assess assumptions. We’re not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot.\n\n\nR\nPython\n\n\n\n\nplot(glm_bks , which = 4)\n\n\n\n\n\n\n\n\n\n\n\n\n\nExtracting the Cook’s distances from the glm object\n\n\n\n\n\nInstead of using the plot() function, we can also extract the values directly from the glm object. We can use the augment() function to do this and create a lollipop or stem plot:\n\nglm_bks %&gt;% \n  augment() %&gt;%            # get underlying data\n  select(.cooksd) %&gt;%      # select the Cook's d\n  mutate(obs = 1:n()) %&gt;%  # create an index column\n  ggplot(aes(x = obs, y = .cooksd)) +\n  geom_point() +\n  geom_segment(aes(x = obs, y = .cooksd, xend = obs, yend = 0))\n\n\n\n\n\n\n\n\n\n\n\n\nAs always, there are different ways of doing this. Here we extract the Cook’s d values from the glm object and put them in a Pandas DataFrame. We can then use that to plot them in a lollipop or stem plot.\n\n# extract the Cook's distances\nglm_bks_py_resid = pd.DataFrame(glm_bks_py.\n                                get_influence().\n                                summary_frame()[\"cooks_d\"])\n\n# add row index \nglm_bks_py_resid['obs'] = glm_bks_py_resid.reset_index().index\n\nWe now have two columns:\n\nglm_bks_py_resid\n\n         cooks_d  obs\n0   1.854360e-07    0\n1   3.388262e-07    1\n2   3.217960e-05    2\n3   1.194847e-05    3\n4   6.643975e-06    4\n..           ...  ...\n56  1.225519e-05   56\n57  2.484468e-05   57\n58  6.781364e-06   58\n59  1.850240e-07   59\n60  3.532360e-05   60\n\n[61 rows x 2 columns]\n\n\nWe can use these to create the plot:\n\n(ggplot(glm_bks_py_resid,\n         aes(x = \"obs\",\n             y = \"cooks_d\")) +\n     geom_segment(aes(x = \"obs\",y = \"cooks_d\", xend = \"obs\", yend = 0)) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nThis shows that there are no very obvious influential points. You could regard point 34 as potentially influential (it’s got a Cook’s distance of around 0.8), but I’m not overly worried.\nIf we were worried, we’d remove the troublesome data point, re-run the analysis and see if that changes the statistical outcome. If so, then our entire (statistical) conclusion hinges on one data point, which is not a very robust bit of research. If it doesn’t change our significance, then all is well, even though that data point is influential.",
+    "text": "7.6 Assumptions\nAs explained in the background chapter, we can’t really use the standard diagnostic plots to assess assumptions. We’re not going to go into a lot of detail for now, but there is one thing that we can do: look for influential points using the Cook’s distance plot.\n\n\nR\nPython\n\n\n\n\nplot(glm_bks , which = 4)\n\n\n\n\n\n\n\n\n\n\n\n\n\nExtracting the Cook’s distances from the glm object\n\n\n\n\n\nInstead of using the plot() function, we can also extract the values directly from the glm object. We can use the augment() function to do this and create a lollipop or stem plot:\n\nglm_bks %&gt;% \n  augment() %&gt;%            # get underlying data\n  select(.cooksd) %&gt;%      # select the Cook's d\n  mutate(obs = 1:n()) %&gt;%  # create an index column\n  ggplot(aes(x = obs, y = .cooksd)) +\n  geom_point() +\n  geom_segment(aes(x = obs, y = .cooksd, xend = obs, yend = 0))\n\n\n\n\n\n\n\n\n\n\n\n\nAs always, there are different ways of doing this. Here we extract the Cook’s d values from the glm object and put them in a Pandas DataFrame. We can then use that to plot them in a lollipop or stem plot.\n\n# extract the Cook's distances\nglm_bks_py_resid = pd.DataFrame(glm_bks_py.\n                                get_influence().\n                                summary_frame()[\"cooks_d\"])\n\n# add row index \nglm_bks_py_resid['obs'] = glm_bks_py_resid.reset_index().index\n\nWe now have two columns:\n\nglm_bks_py_resid\n\n         cooks_d  obs\n0   1.854360e-07    0\n1   3.388262e-07    1\n2   3.217960e-05    2\n3   1.194847e-05    3\n4   6.643975e-06    4\n..           ...  ...\n56  1.225519e-05   56\n57  2.484468e-05   57\n58  6.781364e-06   58\n59  1.850240e-07   59\n60  3.532360e-05   60\n\n[61 rows x 2 columns]\n\n\nWe can use these to create the plot:\n\n(ggplot(glm_bks_py_resid,\n         aes(x = \"obs\",\n             y = \"cooks_d\")) +\n     geom_segment(aes(x = \"obs\", y = \"cooks_d\", xend = \"obs\", yend = 0)) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nThis shows that there are no very obvious influential points. You could regard point 34 as potentially influential (it’s got a Cook’s distance of around 0.8), but I’m not overly worried.\nIf we were worried, we’d remove the troublesome data point, re-run the analysis and see if that changes the statistical outcome. If so, then our entire (statistical) conclusion hinges on one data point, which is not a very robust bit of research. If it doesn’t change our significance, then all is well, even though that data point is influential.",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -246,7 +246,7 @@
     "href": "materials/glm-practical-logistic-binary.html#assessing-significance",
     "title": "\n7  Binary response\n",
     "section": "\n7.7 Assessing significance",
-    "text": "7.7 Assessing significance\nWe can ask several questions.\nIs the model well-specified?\nRoughly speaking this asks “can our model predict our data reasonably well?”\n\n\nR\nPython\n\n\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves.\n\npchisq(9.1879, 59, lower.tail = FALSE)\n\n[1] 1\n\n\nHere, we’ve used the pchisq function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\n\n\nfrom scipy.stats import chi2\n\nchi2.sf(9.1879, 59)\n\n0.9999999999999916\n\n\n\n\n\nThis gives us a probability of 1. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\nIs the overall model better than the null model?\n\n\nR\nPython\n\n\n\n\npchisq(84.5476 - 9.1879, 60 - 59, lower.tail = FALSE)\n\n[1] 3.923163e-18\n\n\nHere we’ve used the pchisq function again (if you didn’t ask before, you probably aren’t going to ask now).\n\n\nFirst we need to define the null model:\n\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ 1\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_null_py = model.fit()\n\nprint(glm_bks_null_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       60\nModel Family:                Binomial   Df Model:                            0\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -42.274\nDate:                Wed, 31 Jan 2024   Deviance:                       84.548\nTime:                        14:26:35   Pearson chi2:                     61.0\nNo. Iterations:                     3   Pseudo R-squ. (CS):              0.000\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept      0.0328      0.256      0.128      0.898      -0.469       0.535\n==============================================================================\n\n\nIn order to compare our original fitted model to the null model we need to know the deviances of both models and the residual degrees of freedom of both models, which we could get from the summary method.\n\nchi2.sf(84.5476 - 9.1879, 60 - 59)\n\n3.9231627082752525e-18\n\n\n\n\n\nThe first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\nThis gives us a probability of pretty much zero. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with blength as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\nAre any of the individual predictors significant?\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the beak length predictor is significant.\n\n\nR\nPython\n\n\n\n\nanova(glm_bks, test = \"Chisq\")\n\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(&gt;Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 &lt; 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nThe anova() function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\nThe p-value for the blength predictor is written under then Pr(&gt;Chi) column and we can see that it is less than &lt; 2.2e-16. So, beak length is a significant predictor.\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n\nAlas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, not beknownst to me…",
+    "text": "7.7 Assessing significance\nWe can ask several questions.\nIs the model well-specified?\nRoughly speaking this asks “can our model predict our data reasonably well?”\n\n\nR\nPython\n\n\n\nUnfortunately, there isn’t a single command that does this for us, and we have to lift some of the numbers from the summary output ourselves.\n\npchisq(9.1879, 59, lower.tail = FALSE)\n\n[1] 1\n\n\nHere, we’ve used the pchisq function (which calculates the correct probability for us – ask if you want a hand-wavy explanation). The first argument to it is the residual deviance value from the summary table, the second argument to is the residual degrees of freedom argument from the same table.\n\n\n\nfrom scipy.stats import chi2\n\nchi2.sf(9.1879, 59)\n\n0.9999999999999916\n\n\n\n\n\nThis gives us a probability of 1. We can interpret this as the probability that the model is actually good. There aren’t any strict conventions on how to interpret this value but, for me, a tiny value would indicate a rubbish model.\nIs the overall model better than the null model?\n\n\nR\nPython\n\n\n\n\npchisq(84.5476 - 9.1879, 60 - 59, lower.tail = FALSE)\n\n[1] 3.923163e-18\n\n\nHere we’ve used the pchisq function again (if you didn’t ask before, you probably aren’t going to ask now).\n\n\nFirst we need to define the null model:\n\n# create a linear model\nmodel = smf.glm(formula = \"pointed_beak ~ 1\",\n                family = sm.families.Binomial(),\n                data = early_finches_py)\n# and get the fitted parameters of the model\nglm_bks_null_py = model.fit()\n\nprint(glm_bks_null_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:           pointed_beak   No. Observations:                   61\nModel:                            GLM   Df Residuals:                       60\nModel Family:                Binomial   Df Model:                            0\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -42.274\nDate:                Thu, 01 Feb 2024   Deviance:                       84.548\nTime:                        07:29:43   Pearson chi2:                     61.0\nNo. Iterations:                     3   Pseudo R-squ. (CS):              0.000\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept      0.0328      0.256      0.128      0.898      -0.469       0.535\n==============================================================================\n\n\nIn order to compare our original fitted model to the null model we need to know the deviances of both models and the residual degrees of freedom of both models, which we could get from the summary method.\n\nchi2.sf(84.5476 - 9.1879, 60 - 59)\n\n3.9231627082752525e-18\n\n\n\n\n\nThe first argument is the difference between the null and residual deviances and the second argument is the difference in degrees of freedom between the null and residual models. All of these values can be lifted from the summary table.\nThis gives us a probability of pretty much zero. This value is doing a formal test to see whether our fitted model is significantly different from the null model. Here we can treat this a classical hypothesis test and since this p-value is less than 0.05 then we can say that our fitted model (with blength as a predictor variable) is definitely better than the null model (which has no predictor variables). Woohoo!\nAre any of the individual predictors significant?\nFinally, we’ll use the anova function from before to determine which predictor variables are important, and specifically in this case whether the beak length predictor is significant.\n\n\nR\nPython\n\n\n\n\nanova(glm_bks, test = \"Chisq\")\n\nAnalysis of Deviance Table\n\nModel: binomial, link: logit\n\nResponse: pointed_beak\n\nTerms added sequentially (first to last)\n\n        Df Deviance Resid. Df Resid. Dev  Pr(&gt;Chi)    \nNULL                       60     84.548              \nblength  1    75.36        59      9.188 &lt; 2.2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n\nThe anova() function is a true workhorse within R! This time we’ve used it to create an Analysis of Deviance table. This is exactly equivalent to an ordinary ANOVA table where we have rows corresponding to each predictor variable and a p-value telling us whether that variable is significant or not.\nThe p-value for the blength predictor is written under then Pr(&gt;Chi) column and we can see that it is less than &lt; 2.2e-16. So, beak length is a significant predictor.\nThis shouldn’t be surprising since we already saw that our overall model was better than the null model, which in this case is exactly the same as asking whether the beak length term is significant. However, in more complicated models with multiple predictors these two comparisons (and p-values) won’t be the same.\n\n\nAlas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, unbeknownst to me…",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
@@ -257,7 +257,7 @@
     "href": "materials/glm-practical-logistic-binary.html#exercises",
     "title": "\n7  Binary response\n",
     "section": "\n7.8 Exercises",
-    "text": "7.8 Exercises\n\n7.8.1 Diabetes\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/diabetes.csv.\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called test_result, glucose and diastolic.\nWe want to see if the glucose tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\nLoad and visualise the data\nCreate a suitable model\nDetermine if there are any statistically significant predictors\nCalculate the probability of a positive diabetes test result for a glucose tolerance test value of glucose = 150\n\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\ndiabetes &lt;- read_csv(\"data/diabetes.csv\")\n\n\n\n\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n\n\n\n\nLooking at the data, we can see that the test_result column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data, by outcome:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the test_result data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\nWe can force Python to temporarily covert the data to a factor, by making the test_result column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nCreate a suitable model\n\n\nR\nPython\n\n\n\nWe’ll use the glm() function to create a generalised linear model. Here we save the model in an object called glm_dia:\n\nglm_dia &lt;- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n\nThe format of this function is similar to that used by the lm() function for linear models. The important difference is that we must specify the family of error distribution to use. For logistic regression we must set the family to binomial.\n\n\n\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n\n\n\n\nLet’s look at the model parameters:\n\n\nR\nPython\n\n\n\n\nsummary(glm_dia)\n\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) -5.611732   0.442289  -12.69   &lt;2e-16 ***\nglucose      0.039510   0.003398   11.63   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n\n\n\nprint(glm_dia_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Wed, 31 Jan 2024   Deviance:                       752.20\nTime:                        14:26:37   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n\n\n\n\n\nWe can see that glucose is a significant predictor for the test_result (the \\(p\\) value is much smaller than 0.05).\nKnowing this, we’re interested in the coefficients. We have an intercept of -5.61 and 0.0395 for glucose. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\\[ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} \\]\nCalculating probabilities\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given glucose value. If we do this for glucose = 150, we get the following:\n\n\nR\nPython\n\n\n\n\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n\n[1] 0.5962827\n\n\n\n\n\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n\n0.5962826992967878\n\n\n\n\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.",
+    "text": "7.8 Exercises\n\n7.8.1 Diabetes\n\n\n\n\n\n\nExercise\n\n\n\n\n\n\n\nLevel: \nFor this exercise we’ll be using the data from data/diabetes.csv.\nThis is a data set comprising 768 observations of three variables (one dependent and two predictor variables). This records the results of a diabetes test result as a binary variable (1 is a positive result, 0 is a negative result), along with the result of a glucose tolerance test and the diastolic blood pressure for each of 768 women. The variables are called test_result, glucose and diastolic.\nWe want to see if the glucose tolerance is a meaningful predictor for predictions on a diabetes test. To investigate this, do the following:\n\nLoad and visualise the data\nCreate a suitable model\nDetermine if there are any statistically significant predictors\nCalculate the probability of a positive diabetes test result for a glucose tolerance test value of glucose = 150\n\n\n\n\n\n\n\n\nAnswer\n\n\n\n\n\n\n\nLoad and visualise the data\nFirst we load the data, then we visualise it.\n\n\nR\nPython\n\n\n\n\ndiabetes &lt;- read_csv(\"data/diabetes.csv\")\n\n\n\n\ndiabetes_py = pd.read_csv(\"data/diabetes.csv\")\n\n\n\n\nLooking at the data, we can see that the test_result column contains zeros and ones. These are yes/no test result outcomes and not actually numeric representations.\nWe’ll have to deal with this soon. For now, we can plot the data, by outcome:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = factor(test_result),\n           y = glucose)) +\n  geom_boxplot()\n\n\n\n\n\n\n\n\n\nWe could just give Python the test_result data directly, but then it would view the values as numeric. Which doesn’t really work, because we have two groups as such: those with a negative diabetes test result, and those with a positive one.\nWe can force Python to temporarily covert the data to a factor, by making the test_result column an object type. We can do this directly inside the ggplot() function.\n\n(ggplot(diabetes_py,\n         aes(x = diabetes_py.test_result.astype(object),\n             y = \"glucose\")) +\n     geom_boxplot())\n\n\n\n\n\n\n\n\n\n\nIt looks as though the patients with a positive diabetes test have slightly higher glucose levels than those with a negative diabetes test.\nWe can visualise that differently by plotting all the data points as a classic binary response plot:\n\n\nR\nPython\n\n\n\n\nggplot(diabetes,\n       aes(x = glucose,\n           y = test_result)) +\n  geom_point()\n\n\n\n\n\n\n\n\n\n\n(ggplot(diabetes_py,\n         aes(x = \"glucose\",\n             y = \"test_result\")) +\n     geom_point())\n\n\n\n\n\n\n\n\n\n\nCreate a suitable model\n\n\nR\nPython\n\n\n\nWe’ll use the glm() function to create a generalised linear model. Here we save the model in an object called glm_dia:\n\nglm_dia &lt;- glm(test_result ~ glucose,\n               family = binomial,\n               data = diabetes)\n\nThe format of this function is similar to that used by the lm() function for linear models. The important difference is that we must specify the family of error distribution to use. For logistic regression we must set the family to binomial.\n\n\n\n# create a linear model\nmodel = smf.glm(formula = \"test_result ~ glucose\",\n                family = sm.families.Binomial(),\n                data = diabetes_py)\n# and get the fitted parameters of the model\nglm_dia_py = model.fit()\n\n\n\n\nLet’s look at the model parameters:\n\n\nR\nPython\n\n\n\n\nsummary(glm_dia)\n\n\nCall:\nglm(formula = test_result ~ glucose, family = binomial, data = diabetes)\n\nCoefficients:\n             Estimate Std. Error z value Pr(&gt;|z|)    \n(Intercept) -5.611732   0.442289  -12.69   &lt;2e-16 ***\nglucose      0.039510   0.003398   11.63   &lt;2e-16 ***\n---\nSignif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\n    Null deviance: 936.6  on 727  degrees of freedom\nResidual deviance: 752.2  on 726  degrees of freedom\nAIC: 756.2\n\nNumber of Fisher Scoring iterations: 4\n\n\n\n\n\nprint(glm_dia_py.summary())\n\n                 Generalized Linear Model Regression Results                  \n==============================================================================\nDep. Variable:            test_result   No. Observations:                  728\nModel:                            GLM   Df Residuals:                      726\nModel Family:                Binomial   Df Model:                            1\nLink Function:                  Logit   Scale:                          1.0000\nMethod:                          IRLS   Log-Likelihood:                -376.10\nDate:                Thu, 01 Feb 2024   Deviance:                       752.20\nTime:                        07:29:45   Pearson chi2:                     713.\nNo. Iterations:                     4   Pseudo R-squ. (CS):             0.2238\nCovariance Type:            nonrobust                                         \n==============================================================================\n                 coef    std err          z      P&gt;|z|      [0.025      0.975]\n------------------------------------------------------------------------------\nIntercept     -5.6117      0.442    -12.688      0.000      -6.479      -4.745\nglucose        0.0395      0.003     11.628      0.000       0.033       0.046\n==============================================================================\n\n\n\n\n\nWe can see that glucose is a significant predictor for the test_result (the \\(p\\) value is much smaller than 0.05).\nKnowing this, we’re interested in the coefficients. We have an intercept of -5.61 and 0.0395 for glucose. We can use these coefficients to write a formula that describes the potential relationship between the probability of having a positive test result, dependent on the glucose tolerance level value:\n\\[ P(positive \\ test\\ result) = \\frac{\\exp(-5.61 + 0.04 \\times glucose)}{1 + \\exp(-5.61 + 0.04 \\times glucose)} \\]\nCalculating probabilities\nUsing the formula above, we can now calculate the probability of having a positive test result, for a given glucose value. If we do this for glucose = 150, we get the following:\n\n\nR\nPython\n\n\n\n\nexp(-5.61 + 0.04 * 150) / (1 + exp(-5.61 + 0.04 * 150))\n\n[1] 0.5962827\n\n\n\n\n\nmath.exp(-5.61 + 0.04 * 150) / (1 + math.exp(-5.61 + 0.04 * 150))\n\n0.5962826992967878\n\n\n\n\n\nThis tells us that the probability of having a positive diabetes test result, given a glucose tolerance level of 150 is around 60 %.",
     "crumbs": [
       "Binary and proportional data",
       "<span class='chapter-number'>7</span>  <span class='chapter-title'>Binary response</span>"
diff --git a/materials/glm-practical-logistic-binary.qmd b/materials/glm-practical-logistic-binary.qmd
index 6338570..c80b463 100644
--- a/materials/glm-practical-logistic-binary.qmd
+++ b/materials/glm-practical-logistic-binary.qmd
@@ -217,11 +217,14 @@ lm_bks_py = model.fit()
 Next, we can create the diagnostic plots:
 ```{python}
 #| eval: false
+#| results: hide
 dgplots(lm_bks_py)
 ```
 
 ```{python}
 #| echo: false
+#| fig-keep: none
+#| fig-show: hide
 # load dgplots function for knitr
 exec(open('setup_files/dgplots_knitr.py').read())
 # create rendered diagnostic plots image
@@ -368,7 +371,6 @@ There’s a lot to unpack here, but let's start with what we're familiar with: c
 ## Parameter interpretation
 
 ::: {.panel-tabset group="language"}
-
 ## R
 
 The coefficients or parameters can be found in the `Coefficients` block. The main numbers to extract from the output are the two numbers underneath `Estimate.Std`:
@@ -574,7 +576,7 @@ We can use these to create the plot:
 (ggplot(glm_bks_py_resid,
          aes(x = "obs",
              y = "cooks_d")) +
-     geom_segment(aes(x = "obs",y = "cooks_d", xend = "obs", yend = 0)) +
+     geom_segment(aes(x = "obs", y = "cooks_d", xend = "obs", yend = 0)) +
      geom_point())
 ```
 
@@ -673,13 +675,11 @@ This shouldn’t be surprising since we already saw that our overall model was b
 
 ## Python
 
-Alas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, not beknownst to me...
+Alas, for some inexplicable reason this is not (yet?) possible to do in Python. At least, unbeknownst to me...
 
 :::
 
 
-
-
 ## Exercises
 
 ### Diabetes {#sec-exr_diabetes}
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